Finding Standard Deviation from a Frequency Distribution. In Exercises 37-40, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4. Standard deviation for frequency distribution s = n [ Σ ( f ⋅ x 2 ) ] − [ Σ ( f ⋅ x ) ] 2 n ( n − 1 ) 40.
Finding Standard Deviation from a Frequency Distribution. In Exercises 37-40, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4. Standard deviation for frequency distribution s = n [ Σ ( f ⋅ x 2 ) ] − [ Σ ( f ⋅ x ) ] 2 n ( n − 1 ) 40.
Solution Summary: The author explains that the standard deviation for the frequency distribution of blood platelet count of females is 69.5.
Finding Standard Deviation from a Frequency Distribution. In Exercises 37-40, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4.
Standard deviation for frequency distribution
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Please could you explain why 0.5 was added to each upper limpit of the intervals.Thanks
28. (a) Under what conditions do we say that two random variables X and Y are
independent?
(b) Demonstrate that if X and Y are independent, then it follows that E(XY) =
E(X)E(Y);
(e) Show by a counter example that the converse of (ii) is not necessarily true.
1. Let X and Y be random variables and suppose that A = F. Prove that
Z XI(A)+YI(A) is a random variable.
Chapter 3 Solutions
Essentials of Statistics, Books a la Carte Edition (6th Edition)
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